1.

If `.^(n)C_(8)=.^(n)C_(6)`, then find `.^(n)C_(2)`.

Answer» If `.^(n)C_(x)= .^(n)C_(y) " and" x ne y`, then x+y=n. Hence,
`.^(n)C_(8)= .^(n)C_(6)`
`implies n=(8+6)=14`
Now, `.^(n)C_(2)= .^(14)C_(2)=(14xx13)/(2)=91`


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